0.00/0.75
0.00/0.75
(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(empty, l) → l
0.00/0.75
f(cons(x, k), l) → g(k, l, cons(x, k))
0.00/0.75
g(a, b, c) → f(a, cons(b, c))
Rewrite Strategy: INNERMOST
0.00/0.75
0.00/0.75
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
0.00/0.75
0.00/0.75
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(empty, z0) → z0
0.00/0.75
f(cons(z0, z1), z2) → g(z1, z2, cons(z0, z1))
0.00/0.75
g(z0, z1, z2) → f(z0, cons(z1, z2))
Tuples:
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
S tuples:
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
0.00/0.75
0.00/0.75
(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
We considered the (Usable) Rules:none
And the Tuples:
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :
0.00/0.75
POL(F(x1, x2)) = [4]x1
0.00/0.75
POL(G(x1, x2, x3)) = [1] + [4]x1
0.00/0.75
POL(c1(x1)) = x1
0.00/0.75
POL(c2(x1)) = x1
0.00/0.75
POL(cons(x1, x2)) = [4] + x2
0.00/0.75
0.00/0.75
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(empty, z0) → z0
0.00/0.75
f(cons(z0, z1), z2) → g(z1, z2, cons(z0, z1))
0.00/0.75
g(z0, z1, z2) → f(z0, cons(z1, z2))
Tuples:
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
S tuples:none
K tuples:
F(cons(z0, z1), z2) → c1(G(z1, z2, cons(z0, z1)))
0.00/0.75
G(z0, z1, z2) → c2(F(z0, cons(z1, z2)))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
0.00/0.75
0.00/0.75
(5) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
0.00/0.75
0.00/0.75
(6) BOUNDS(O(1), O(1))
0.00/0.75